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A 45 kg crate accelerates at 1.65 m/s2 when pulled by a rope with a force of 200 N. Find the angle the rope is pulled at. Friction is negligible.
68.2° above the horizon
| Step | Formula / Calculation | Reasoning |
|---|---|---|
| 1 | F_{\text{net}} = m \cdot a | Net force calculation using Newton’s second law |
| 2 | F_{\text{horizontal}} = F_{\text{net}} | Horizontal force component causing acceleration |
| 3 | F_{\text{vertical}} = \sqrt{F^2 – F_{\text{horizontal}}^2} | Vertical force component calculation |
| 4 | \tan(\theta) = \frac{F_{\text{vertical}}}{F_{\text{horizontal}}} | Relationship between force components and angle |
| 5 | \theta = \arctan\left(\frac{F_{\text{vertical}}}{F_{\text{horizontal}}}\right) | Calculating the angle in radians |
| 6 | \theta_{\text{degrees}} = \frac{\theta \cdot 180}{\pi} | Converting angle to degrees |
| 7 | 68.21° | Final Answer |
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Note: This is a bonus question. Skip if you haven’t yet taken calculus.
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Bonus Challenge: Repeat the problem but assume you are not given the initial speed, angle of incline, or µk.
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68.2° above the horizon
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| Kinematics | Forces |
|---|---|
| \Delta x = v_i \cdot t + \frac{1}{2} a \cdot t^2 | F = m \cdot a |
| v = v_i + a \cdot t | F_g = \frac{G \cdot m_1 \cdot m_2}{r^2} |
| a = \frac{\Delta v}{\Delta t} | f = \mu \cdot N |
| R = \frac{v_i^2 \cdot \sin(2\theta)}{g} |
| Circular Motion | Energy |
|---|---|
| F_c = \frac{m \cdot v^2}{r} | KE = \frac{1}{2} m \cdot v^2 |
| a_c = \frac{v^2}{r} | PE = m \cdot g \cdot h |
| KE_i + PE_i = KE_f + PE_f |
| Momentum | Torque and Rotations |
|---|---|
| p = m \cdot v | \tau = r \cdot F \cdot \sin(\theta) |
| J = \Delta p | I = \sum m \cdot r^2 |
| p_i = p_f | L = I \cdot \omega |
| Simple Harmonic Motion |
|---|
| F = -k \cdot x |
| T = 2\pi \sqrt{\frac{l}{g}} |
| T = 2\pi \sqrt{\frac{m}{k}} |
| Constant | Description |
|---|---|
| g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
| G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
| \mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |
| k | Spring constant, in \text{N/m} |
| Variable | SI Unit |
|---|---|
| s (Displacement) | \text{meters (m)} |
| v (Velocity) | \text{meters per second (m/s)} |
| a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |
| t (Time) | \text{seconds (s)} |
| m (Mass) | \text{kilograms (kg)} |
| Variable | Derived SI Unit |
|---|---|
| F (Force) | \text{newtons (N)} |
| E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |
| P (Power) | \text{watts (W)} |
| p (Momentum) | \text{kilogram meters per second (kg·m/s)} |
| \omega (Angular Velocity) | \text{radians per second (rad/s)} |
| \tau (Torque) | \text{newton meters (N·m)} |
| I (Moment of Inertia) | \text{kilogram meter squared (kg·m}^2\text{)} |
| f (Frequency) | \text{hertz (Hz)} |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: \text{5 km}
Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}
Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | 10^{-12} | |
Nano- | n | 10^{-9} | |
Micro- | µ | 10^{-6} | |
Milli- | m | 10^{-3} | |
Centi- | c | 10^{-2} | |
Deci- | d | 10^{-1} | |
(Base unit) | – | 10^{0} | |
Deca- or Deka- | da | 10^{1} | |
Hecto- | h | 10^{2} | |
Kilo- | k | 10^{3} | |
Mega- | M | 10^{6} | |
Giga- | G | 10^{9} | |
Tera- | T | 10^{12} |
